Question: Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{-2})^{-5}}}{{(y^{4}r^{-2})^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{-2}}$ to the exponent ${-5}$ . Now ${-2 \times -5 = 10}$ , so ${(y^{-2})^{-5} = y^{10}}$ In the denominator, we can use the distributive property of exponents. ${(y^{4}r^{-2})^{3} = (y^{4})^{3}(r^{-2})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{-2})^{-5}}}{{(y^{4}r^{-2})^{3}}} = \dfrac{{y^{10}}}{{y^{12}r^{-6}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{10}}}{{y^{12}r^{-6}}} = \dfrac{{y^{10}}}{{y^{12}}} \cdot \dfrac{{1}}{{r^{-6}}} = y^{{10} - {12}} \cdot r^{- {(-6)}} = y^{-2}r^{6}$.